# Solution to the Circle in a Hexagon in a Semi-Circle Puzzle +-- {.image} [[CircleinaHexagoninaSemiCircle.png:pic]] > The hexagon is regular. What fraction of the semicircle is shaded? =-- ## Solution by [[Lengths in a Regular Hexagon]] and [[Pythagoras' Theorem]] +-- {.image} [[CircleinaHexagoninaSemiCircleLabelled.png:pic]] =-- In the above diagram, the point $O$ is the centre of the circle and the point $A$ is the midpoint of the top edge of the hexagon. As that top edge is a [[chord]] of the circle, its [[perpendicular bisector]] passes through the centre so $O$ is also the midpoint of the bottom edge. Let $r$ be the radius of the shaded circle and $R$ the radius of the semi-circle, so $O A$ has length $2 r$ and $O B$ has length $R$. By considering the [[lengths in a regular hexagon]], the length of $O A$ is $\sqrt{3}$ times the side length of the hexagon, so $A B$ has length $\frac{r}{\sqrt{3}}$. Applying [[Pythagoras' theorem]] to triangle $O A B$ shows that: $$ R^2 = (2 r)^2 + \left( \frac{r}{\sqrt{3}} \right)^2 = 4 r^2 + \frac{r^2}{3} = \frac{13 r^2}{3} $$ Since the area of the circle is $\pi r^2$ and of the semi-circle is $\frac{1}{2} \pi R^2$, the fraction of the semi-circle that is shaded is $\frac{6}{13}$.